In: Physics
A soft drink (mostly water) flows in a pipe at a beverage plant with a mass flow rate that would fill 220 0.355 - L cans per minute. At point 2 in the pipe, the gauge pressure is 152kPa and the cross-sectional area is 8.00cm2. At point 1, 1.35m above point 2, the cross-sectional area is 2.00cm2.
Part a
Find the mass flow rate. kg/s
part b
Find the volume flow rate. L/s
part c
Find the flow speed at point 1. m/s
Mass flow rate \(=\) density \(^{*}\) volume flow rate
now first we will find the volume flow rate
Volume rate\(=\frac{\text {Volume}}{\text {time}}\)
Volume rate \(=\frac{220 * 0.355}{60}\)
PART B)
Volume rate \(=1.30 L / s\)
PART A)
Mass flow rate is given as
Mass rate \(=\) density \(*\) Volume rate
Mass rate \(=1000 * 1.30 * 10^{-3} \mathrm{~m}^{3} / \mathrm{s}\)
Mass rate \(=1.30 \mathrm{~kg} / \mathrm{s}\)
PART C)
by Equation of continuity we can say
\(A_{1} v_{1}=A_{2} v_{2}\)
\(8 * v_{2}=2 * v_{1}=\) flow rate
\(2 * 10^{-4} m^{2} * v_{1}=1.30 * 10^{-3} \mathrm{~m}^{3} / \mathrm{s}\)
\(v_{1}=\frac{1.30 * 10^{-3}}{2 * 10^{-4}}\)
\(v_{1}=6.5 \mathrm{~m} / \mathrm{s}\)